Charge given externally to a capacitor

[EddiePhys]

Homework Statement

A capacitor of capacitance C is charged to a potential difference v from the cell and then disconnected from it. A charge Q is now given to its positive plate. The potential difference across the capacitor is now?

Homework Equations

The Attempt at a Solution

V is the potential difference. Charge +Q is given and hence -Q is induced on the negative plate.

The potential difference should be V+Q/C
But the answer is V+Q/2C. Why?

 

[NFuller]

Charge +Q is given and hence -Q is induced on the negative plate.

There is no -Q induced on the negative plate since according to the problem the capacitor has been disconnected from the circuit; there is no path for charge to move onto or off of the negative plate. So there is only a +Q charge on the positive plate.

The potential difference should be V+Q/C
But the answer is V+Q/2C. Why?

The formula $V=Q/C$ assumes equal and opposite charge on both plates. Since only one plate was charged here, it is only half of this value $Q/2C$.

 

[rude man]

You can solve this formally as follows:
Assign surface charge densities σ₁, σ₂, -σ₂ and σ₃ to the two plates. The surfaces 2 and -2 face each other inside the capacitor. Make sure you understand why the inside surface charge densities have to be equal and opposite. Assume unity area. then you can write 3 equations in 3 unknowns and solve for all three surface charge densities.
Two of the equations are simply charge summation. For the third I'll give you a hint: what must the 3rd equation be in order to make the E field zero inside either plate?
Once you have surface charge densities inside the capacitor it should be straightforward to find the E field and integrate, with and without the extra charge.

PS the given answer is what I got.